Real Analysis 14 | Heine-Borel Theorem [dark version]

Heine–Borel theorem for real numbers boils compactness down to two checkable conditions: a set is compact exactly when it is both bounded and closed. That matters because “compact” is defined through behavior of every sequence—every sequence drawn from the set must have an accumulation point that still lies in the set—while boundedness and closedness are often easier to verify directly.

The discussion starts by recalling a sequential definition of compactness: a subset A of ℝ is compact if every sequence whose terms lie in A has at least one accumulation (limit) value, and crucially that accumulation value belongs to A. Simple examples follow. The empty set is compact because there are no sequences to test. A one-point set is compact since the only relevant sequence is constant, so its accumulation value is that same point.

To show what fails, the transcript considers an “unbounded” example: take a sequence that increases without bound (diverging to infinity). Such a sequence has no accumulation value in ℝ, so the set containing its terms cannot be compact. This sets up the need for boundedness.

Next comes the key example of a closed interval [C, D]. Any sequence inside [C, D] is automatically bounded. Bolzano–Weierstrass then guarantees that every bounded sequence in ℝ has an accumulation value. The remaining question is whether that accumulation value stays inside the interval; closedness answers that. Because [C, D] is closed, limits of sequences drawn from it cannot “escape,” so the accumulation point lies in [C, D]. That combination—boundedness plus closedness—yields compactness for closed intervals.

With those ingredients in place, the Heine–Borel theorem is stated: for any subset A ⊆ ℝ, A is compact if and only if A is bounded and closed. One direction is already essentially built from the interval argument. If A is bounded, Bolzano–Weierstrass provides an accumulation value for any sequence in A. If A is also closed, the accumulation value must lie in A, matching the sequential definition of compactness.

For the other direction, compactness forces both properties. To prove closedness, take any convergent sequence in A with limit ã. Compactness implies the sequence has an accumulation value that must lie in A. Since a convergent sequence has only one accumulation value, that accumulation value equals the limit ã, so ã ∈ A; therefore A is closed.

To prove boundedness, the transcript uses proof by contradiction: assume A is not bounded. Then one can construct a sequence (a_n) in A with |a_n| > n for every natural number n. Such a sequence cannot have any accumulation value in ℝ, contradicting compactness. Hence A must be bounded.

The result is presented as a practical replacement: in ℝ, verifying compactness becomes verifying boundedness and closedness together—no need to check accumulation points for every sequence.